Analytic functions (part 1)

I am obligated to mention that I made a mistake when talking about holomorphic functions. To check that ${f}$ is holomorphic, you must check that ${f}$ is continuous and the partials ${u_x}$, etc exist and satisfy the Cauchy-Riemann equations. I forgot to mention that ${f}$ needs to be continuous. The proof relies on the continuity, and I even used this fact in the examples.

Now let’s talk about analytic functions and power series. Let ${U}$ be an open set in ${{\mathbb C}}$, and let ${f:U\rightarrow{\mathbb C}}$. Then we say ${f}$ is analytic if locally, it is given by a power series. This is to say, it doesn’t have to have a unique power series representation on all of ${U}$, but if you pick a point ${z_0\in U}$, there’s a neighborhood around ${z_0}$ for which ${f}$ is given by a power series centered at ${z_0}$. By power series centered at ${z_0}$, I mean that, in this small neighborhood of ${z_0}$,

$\displaystyle f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots,$

for ${a_n\in{\mathbb C}}$. Don’t forget that even though we can pretend, we really don’t add up infinitely many terms. What we have really done is defined the finite sum ${f_n(z)=a_0+a_1(z-z_0)+\cdots+a_n(z-z_0)^n}$, and let

$\displaystyle f(z)=\displaystyle\lim_{n\rightarrow\infty}f_n(z).$

The distinction is important.

But these power series don’t always converge. In the case of real power series, recall that ${1/(1-x)=1+x+x^2+x^3+\cdots}$, but this only makes sense where the right-hand side converges (between ${1}$ and ${-1}$). A similar thing happens over ${{\mathbb C}}$.

For a power series like the one given above, define

$\displaystyle R=\left(\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}.$

The astute observer will recognize that this could be ${1/0}$, or ${1/\infty}$. In such a case, we just say that ${R=\infty}$ or ${R=0}$ respectively. We call ${R}$ the radius of convergence for reasons that the next theorem should make clear. For example, ${R=\infty}$ will mean the power series converges everywhere. There are a few points to note here. If the power series is finite, then for ${n}$ large enough, ${a_n=0}$, meaning ${R=\infty}$. This makes sense, because polynomials are well defined on all of ${{\mathbb C}}$. Also, recognize that this definition is perfectly valid for real power series. In particular, for ${1+x+x^2+\cdots}$, the radius of convergence is 1.

Lemma 1 (Root Test) Let ${f(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots}$ be a power series, and define

$\displaystyle R=\left(\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}.$

Then if ${|z-z_0|, ${f(z)}$ converges. If ${|z-z_0|>R}$, ${f(z)}$ diverges. We make no claims about the case ${|z-z_0|=R}$.

Proof: We need to treat the cases ${R=0}$ and ${R=\infty}$ separately. I won’t bother doing them at all, because they’re really much easier than the case where ${R\in(0,\infty)}$. Suppose first that ${r=|z-z_0|>R}$. Then ${\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}=1/R>1/r}$, meaning there are infinitely many ${n}$ for which ${\sqrt[n]{|a_n|}>1/r}$. That is, there are infinitely many ${n}$ for which

$\displaystyle |a_n|\cdot|z-z_0|^n=|a_n|r^n>1.$

Thus, the terms in the sum ${\sum_{n=0}^\infty a_n(z-z_0)^n}$ do not converge to zero, so there is no way the sum could converge.

On the other hand, if ${r=|z-z_0|, then ${1/r>\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}}$. From the definition of ${\limsup}$, this means there is some point after which all entries are less than ${1/r}$. More technically, there is some number ${N}$ such that for each ${n>N}$,

$\displaystyle \sqrt[n]{|a_n|}<1/r.$

We can even do a bit better than that. Just being less than ${1/r}$, they could get arbitrarily close to ${1/r}$. Of course, if that were the case, then ${1/r}$ would be the ${\limsup}$, which it’s not (it’s greater than the ${\limsup}$). So in fact there is some real number ${x<1/r}$ for for which, whenever ${n>N}$,

$\displaystyle \sqrt[n]{|a_n|}

Now, it follows that ${|a_n|r^n. Since ${x\cdot r<1}$, we have shown that the terms in the power series (after a certain point) grow slower in magnitude than a geometric series of radius less than one. Thus, ${f(z)}$ must converge.

$\Box$

This was a proof of the root test, and a first step towards what we’ll prove next time: That analytic functions are holomorphic.